The proof, with a missing reason, proves that the measure of angle ECB is 54°.

Mathematics

1 answer:

aniked [119]3 years ago

3 0

Answer:

Midsegment of a Triangle Theorem

Step-by-step explanation:

$In\: \triangle ADE\\m\angle A + m\angle D + m\angle AED = 180°\\\therefore 90°+36°+m\angle AED= 180°\\\therefore 126° +m\angle AED = 180°\\\therefore m\angle AED = 180°-126°\\\therefore m\angle AED = 54°\\$

$In\: \triangle ADE,$ D and E are mid points of sides AB and AC respectively.

Therefore, DE || BC

Hence, by Midsegment of a Triangle Theorem

$m\angle ACB = m\angle AED =54°(corresponding\:\angle 's) \\\therefore m\angle ECB = m\angle AED =54°(\because A - E - C)$

You might be interested in

(a) Show that a differentiable function f decreases most rapidly at x in the direction opposite the gradient vector, that is, in

Sophie [7]

Answer:

Step-by-step explanation:

$\text{Show that a differentiable function f decreases most rapidly at x in the }$

$\text{direction opposite the gradient vector, that is, in the direction of}$ $-\bigtriangledown f(x)$ $\text{. Let}\ \theta \ \text{be the angle between} \bigtriangledown f(x) \ \text{and unit vector u. Then } D_u f = \mathbf{|\bigtriangledown f| \ cos \ \theta }}$

$\text{Since the minimum value of} \ \ \mathbf{cos \ \theta} \ \ is \mathbf{-1} \ \text{occuring \ for \ 0} \le \ \theta \ < 2x, \\ \\ when \ \theta = \mathbf{\pi} , \text{the mnimum value of} \ D_uf \ is} -|\bigtriangledown f|, \text{occuring when the direction of u is } \\ \\ \ \mathbf{the \ opposite \ of} \ \text{the direction of } \ \bigtriangledown f (assuming \ \bigtriangledown f\ is \ not \ zero)$

b) $\text{From part A:}$

$If \ f(x,y) = x^4y -x^2y^2 \ \ decreases \ fastest \ at \ the \point \ (2,-5)\\ \\ F(x,y) = x^4y -x^2y^3 \\ \\ f_x = \dfrac{df}{dx}= \dfrac{d}{dx}(x^4y-x^2y^3) \\ \\ f_x = \dfrac{df}{dx}= y4x^3 -2y^3x \\ \\ For(2,-5) \\ \\ f_x = (-5)4(2)^3 -2(-5)^3(2) \\ \\ \mathbf{ f_x = 340}$

$However; f_y = \dfrac{df}{dy} = \dfrac{d}{dy}(x^4y - x^2y^3) \\ \\ f_y = x^4 -3x^2y^2 \\ \\ Now, for (2, -5)\\ \\f_y = (2)^4 -3(2)^2(-5)^2 \\ \\ f_y = -284$

$So; \bigtriangledown = < 340,-284> \text{this is the direction of fastest decrease}$

6 0

3 years ago

X^2-5x-8=0 find the roots (zeros)

melomori [17]

Answer:

$x_1=\dfrac{5-\sqrt{57}}{2}\ ,\quad x_2=\dfrac{5+\sqrt{57}}{2}$

Step-by-step explanation:

$x^2-5x-8=0\\\\a=1\,,\ b=-5\,,\ c=-8\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x=\dfrac{-(-5)\pm\sqrt{(-5)^2-4\cdot1\cdot(-8)}}{2\cdot1}=\dfrac{5\pm\sqrt{25+32}}{2}=\dfrac{5\pm\sqrt{57}}{2}$